 Welcome, Myself, Mr. Vishal Angire, Assistant Professor, Department of Electronics and Telecommunication, Wolchen Institute of Technology. Today we will study superposition theorem, which is to be used to solve circuit and network related problems. Learning outcome, at the end of this session, student will be able to apply superposition theorem to find unknown voltage or current through the circuit element. Before studying superposition theorem, you should recall Critchoff's voltage law, Critchoff's current law, series and parallel combination of passive circuit elements, voltage division, current division, star and delta connection and source transformation. Superposition to superposition theorem, so when two or more sources are present in a network and not connected in series or parallel, superposition theorem is very much useful to find circuit parameters like voltage or current. Now we see statement of superposition theorem. If a number of voltage or current sources are acting simultaneously in a linear network, the resultant current in any branch is the algebraic sum of the currents that would be produced in it, when each source acting alone replacing all other independent sources by their internal resistances. So basically in superposition theorem, if we want to find out current through a circuit element, the most condition is sources must be two or more than two and that time we have to analyze effect of each source individually and remaining sources are to be deactivated. So while deactivating the remaining sources, if it is voltage source, we have to repress the voltage source by its internal resistance and if it is current source, we have to delete the current source and then we have to analyze the effect. Now we will see explanation of superposition theorem. So consider the simple DC circuit as shown in this figure and the problem statement is we have to find out current I L flowing through the resistor of 1 kilo ohm and for this we have to use superposition theorem. So while applying superposition to this particular problem, we have to deactivate one voltage source which is of 5 volt and we have to analyze effect of 10 volt voltage source and while analyzing the effect of 10 volt voltage source, we have to consider current flowing through resistor of 1 kilo ohm as a I L 1 and total current is to be reprinted by I. So to find out total current I, we have to use the formula that I is equal to V upon R. V is total voltage and R is total resistance. So total resistance is resistance of 2 kilo 2.2 kilo ohm in series with the parallel combination of 1 kilo ohm resistor and 330 ohm resistor. So current I, total current I is 4.08 milliampere. Next we have to find out current I L 1 flowing through the resistor of 1 kilo ohm for this we have to use current division rule. So formula is I L 1 is equal to total current multiplied by resistance of 330 ohm upon 1k plus 330 ohm. So here current I L 1 is 1.01 milliampere. So this is the current I L 1 flowing through resistor of 1 kilo ohm when we are analyzing the effect of 10 volt voltage source. Next now we have to consider the effect of voltage source 5 volt. So deactivate the voltage source of 10 volt by replacing it through its internal resistance means we have to short circuit the branch. Here again we have to consider total current I flowing through the circuit and current flowing through the resistor of 1 kilo ohm is to be represented by I L 2. Now first we will find total current I which is flowing through the circuit. So again use formula of I is equal to V upon R, I total current V, total voltage and R total resistance. So here current I is equal to 5 volt that is total voltage upon total resistance. So total resistance is resistance of 330 ohm in series with the parallel combination of 2.2 kilo ohm resistance with 1 kilo ohm resistance. So total current I in this case is 4.91 milliampere. Now we have to find out current flowing through the resistor of 1 kilo ohm which is to be represented by I L 2. So here also to find out current I L 2 we have to use current division rule. So formula is I L 2 is equal to total current I into resistor of 2.2 kilo ohm in numerator and in denominator addition of the resistances that is 2.2 kilo ohm and 1 kilo ohm. So current I L 2 flowing through the resistor of 1 kilo ohm when we are analyzing the effect of 5 volt voltage source becomes 3.37 milliampere. So now we got 2 currents that is current I L 1 when we are analyzing the effect of 10 volt voltage source and we are reactivated the voltage source of 5 volt and current I L 2 when we are analyzing the effect of voltage source 5 volt that time we have reactivated the voltage source of 10 volt. Now the total current I L flowing through the resistor of 1 kilo ohm is algebraic sum of currents I L 1 and I L 2 when individual sources are acting alone and remaining sources are to be replaced by their internal resistances. So as per superposition theorem total current I L is equal to I L 1 plus I L 2. So it is 1.01 milliampere plus 3.37 milliampere. So current I L flowing through resistor of 1 kilo ohm becomes 4.38 milliampere. So if you want you can tally the current I L flowing through resistor of 1 kilo ohm by using another method means conventional network solving method and you can verify either the current is same or not. While preparing this video lecture I have referred circuit theorem analysis and synthesis book by A. Chakravarty 6 revised edition thank you.